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PHYSICS 125
Lecture Notes Book 3
Prepared by Kai Wong
Table of Contents
11. Work and Energy ………………………………………………………. 2
12. Conservation of Mechanical Energy ……………………………………9
13. Impulse and Momentum ………………………………………………..15
14. Conservation of Momentum …………………………………………….19
15. Kinematics of Rotation …………………………………………………..30
1
11. Work and Energy

Definition of Work Consider a force F acting on an object that undergoes a

displacement d . There may or may not be other forces acting on the same object, so the

displacement is in general not a direct consequence of the force F . The work done by the
force is thus defined:




Work done by F  Component of F along the direction of d  magnitude of d
Or, in symbols,

W  Fd d



Case 1: F and d are in the same direction. We have Fd  F , and so
 
W Fd
F
d



Case 2: F and d are in opposite directions. We have Fd   F , and so
 
W  F  d
F
d


Case 3: F is perpendicular to d . We have Fd  0 , and so
W 0
2
F
d



Case 4: F makes an angle   90  with d . We have Fd  F cos , and so
 
W  F  d  cos
F

d
Fd




Case 5: F makes an angle   90  with d . We have Fd   F cos 180     F cos  ,




because cos 180      cos  , and so again
 
W  F  d  cos
F
180  

d
Fd
In fact, we can take
 
W  F  d  cos
as the definition of work in general, noting that cos 0   1 cos180  1 cos 90   0
The unit of work is N  m , and is called J (Joule).
Getting a force to do work is often desirable. To move a chair on a rough floor, you
need to push. The purpose of the push force is to overcome friction. The push force does
3
positive work because the chair moves in the direction of the push. Note that the work
done by the friction force in this case is negative, but is no concern usually. We can also
move the chair up by lifting. In this case, the lift force is required to overcome gravity.
Again the lift force does positive work. The negative work done by gravity is usually of
no concern.
Work Energy Theorem

Consider the simple case of an object of mass m being pulled by a single force F . Since
the motion is one dimensional, Newton’s second law gives the acceleration as
a
F
m
Recall the relation between initial velocity, final velocity, and displacement:
v 2  v02  2ax
which we rewrite as
v 2f  vi2  2
F
x
m
Since
W  Fx
we can recast the above equation into the form
W
1 2 1 2
mv f  mvi
2
2
This motivates the definition of kinetic energy as follows:
K
1 2
mv
2
We have established in our simple case of one force that
Work done  Change in kinetic energy
Or, W  K
In general, if there are multiple forces acting on a single object, we need to modify the
above to
4
Sum of Work done by all forces  Change in kinetic energy
W  K
This is called the Work-energy Theorem.
The unit of kinetic energy is also J , the same as for work. Note that kinetic energy is a
scalar, and does not depend on the direction of the velocity vector.
Kinetic energy can be exploited to do work in the following sense. To reduce the
kinetic energy of a moving object to zero, we need to apply a force. The work done by
this force is negative according to work-energy theorem because W  K  0 . This is
also clear when we realize that the direction of this force must be opposing the velocity of
the object to cause it to slow down. The reaction of this force acts on whatever agent that
applies the force in the opposite direction. This reaction is in the same direction as the
displacement, and so does positive work (on the agent). This is the positive work
derivable from the kinetic energy of the object. As a concrete example, consider hitting a
nail with a hammer. The kinetic energy of the hammer disappears when the hammer is
stopped by the nail because the nail exerts a force on the hammer that does negative
work. The force exerted by the hammer is a reaction of this force, and it does positive
work as the nail is driven into the wall.
Displacement of
nail
Displacement of
hammer
Work done by gravity
Consider an object moving from a point A at a height hi to a point B at a height h f as
shown in the diagram. The heights are measured from some reference level. During the
displacement, the gravitational attraction on the object (its weight) is a downward force
equal to mg in magnitude where m is the mass of the object. The work is
Wg  mgd cos   mg hi  h f   mghi  mgh f
Defining the gravitational potential energy of the object at a height h to be
U  mgh
5
we can write
W g   U
A

hi
hi-hf mg
B
hf
Since only the difference in U is involved, the choice of the reference level for height
measurement is immaterial. In particular, h is negative if the object is below the reference
level. In fact, h is a y-coordinate.
According to this relation, we can say that gravity taps into potential energy to do
work, because in moving from a higher elevation to a lower elevation, there is a
reduction of potential energy ( U  0 ), and the work W g   U is positive.
We can show that the work done by gravity in going from one point to the other is
independent of the path followed. To illustrate this, consider taking the path AC and CB
to go from A to B, where the point is directly underneath A and at the same level as B as
shown. Along the path AC, the work is
WAC  mghi  h f cos 0  mghi  h f   U
Along the path BC, the work is
WCB  mgCB cos 90  0
Since WAC  WCB  U , we have
WAB  WAC  WCB
A force for which the work done is independent of path is called a conservative force. If
the path is a closed loop, the work done by a conservative force would be zero, because
6
among all paths starting from a point and ending in the same point, there is the trivial
path involving no movement. For this path, the work is clearly zero.
We have seen that for a conservative force, the work done can be calculated as the
difference between the potential energies at the starting and the end points, without
knowledge of what happens in between.
Friction is not a conservative force. To demonstrate, push a block on a rough table
forward and then backward to return to the same location. During both the forward and
the backward movements, the friction force is in the opposite direction of the
displacement, and does negative work in cases. The net work done by friction for the
closed path is not zero.
Example: A 0.05-kg block slides with constant speed down an incline plane, changing its
elevation from 0.8m above the floor to 0.2m. Find the work done by all the forces on the
block.
Solution: The forces on the object consist of its weight, the normal force, and kinetic
friction.
FK
FN
mg
0.8m
0.2m
The work done by gravity is
Wg  U  mgh  mg h f  hi   mg hi  h f   0.05  10  0.8  0.2  0.3J
The work done by the normal force is
WFN  0
7
because the force is perpendicular to the displacement.
To find the work done by the friction force, we use the work energy theorem, noting that
there is no change in kinetic energy K  0 as the object moves with constant speed:
W g  WFN  WFK  0
WFK  0.3J
8
12 Conservation of Mechanical Energy
Consider an object of mass m moving entirely under the influence of gravity such as a
projectile. The work energy theorem reduces to
W g  K
and leads to
 U  K
Or, K  U  K  U   0
Alternatively, we can write
Ki  Ui  K f  U f
This states that the sum of kinetic and gravitational potential energy remains unchanged
in the course of motion. Kinetic and potential energy are collectively known as
mechanical energy. We have just demonstrated the law of conversation of mechanical
energy.
If we consider
Gain in kinetic energy  K f  K i  K
Loss in potential energy  U i  U f  U
we can interprete conservation of energy as saying that the gain in kinetic energy is equal
to the loss of potential energy. Gains and losses can be either positive and negative.
In the course of motion of the projectile, there is a transformation of mechanical
energy from kinetic to potential and vice versa taking place continually without the total
energy being affected.
Application of this law simplifies the solution of some projectile problems as the
following example shows.
Example: A rock is projected with speed 5m/s on top of a building of height 8m. Find its
speed when it lands.
9
5m/s
i
8m
f
v
Appling conservation of mechanical energy to the initial and final locations as indicated:
1 2
1
mvi  mghi  mv 2f  mgh f
2
2
we see that m cancels out:
1 2
1
vi  ghi  v 2f  gh f
2
2
Putting in values:
1 2
1
 5  10  8  v 2  0
2
2
 v  52  2 10  8  185  13.6m / s
Note that this result is independent of the angle of elevation with which the rock is
launched.
For motion of an object on a frictionless railing such as a roller coaster, there is normal
force in addition to gravitational force. The work energy theorem states that
W g  WFN  K
Since the work done by the normal force W FN is zero, we still expect conservation of
mechanical energy.
10
Example: What is the minimum speed of a car at the bottom (point A) of a roller coaster
in order to reach a peak location (B) 12m above A? What is its speed when it reaches the
location C which is 8m above A but on the other side of the peak ?
Solution:
B
C
12m
8m
A
vA
From conservation of mechanical energy:
1 2
1
v A  gh A  v B2  ghB
2
2
To barely reach the point B, the condition is vB  0 . Since hA  0 by choice, we obtain
1 2
v A  ghB
2
 v A  2 ghB  2 10 12  15.5m / s
Applying conservation of energy between B and C:
1 2
1
v B  ghB  vC2  ghC
2
2
 vC  2 g hB  hC   2  10  4  8.9m / s
If the railing has friction, the law of conservation of mechanical energy breaks down. We
need to include work done by friction, which as we have seen, is a kind of nonconservative force. In general, when non-conservative force is present, we need to
explicitly isolate it in the work energy theorem, which now reads
11
W g  W NC  K
and leads to
K  U   WNC
which is called the generalized work energy theorem
When kinetic friction is the non-conservative force, the work W NC is always negative
because the direction of kinetic friction is opposite to displacement. In this case, the total
mechanical energy K  U always decreases. This phenomenon is known as dissipation.
As a result of dissipation, a steel ball rolling back and forth about the bottom of a railing
eventually comes to a stop. A simple pendulum swinging back and forth in air will also
eventually stop.
Example: A 0.4-kg steel ball is released with zero speed at a point of a railing that is 5m
above the bottom. Its speed is found to be 8m/s at the bottom. What is the work done by
friction?
5m
8m/s
K i  U i  U i  mghi  0.4  10  5  20 J
1
1
K f  U f  K f  mv 2f   0.4  8 2  12.8 J
2
2
WF  K f  U f   K i  U i   12.8  20  7.2 J
If the length of the path followed by the steel is a quarter of a circle, we can find the
friction force itself if it can be assumed to be constant. In this case, the path length is
s
1
 5
 2r  
 7.9m
4
2
Using
12
WF   FK s
we find
FK  
WF 7.2

 0.91N
s
7.9
Power
The average power delivered by a force to an object undergoing a displacement during a
time interval t is defined as:
P
W
t
where W is the work done during the displacement. The unit of work is J/s, and is
abbreviated as W (Watt). The British unit of power is horse power (hp).
1 hp=746W
For an object moving in one dimension and a force F acting in the same dimension, if the
displacement during the time interval t is x , the work done by the force is W  Fx
P 
Fx
 Fv
t
where v is the average velocity.
This formula has an interesting application to the motion of cars. If a car is moving with
constant speed on a level road, the static friction force FS on its tires (traction) is forward
and balances air resistance FR. The power delivered by traction is therefore
P  FS v  FR v
The power increases with the speed of the car both because of its explicit dependence on
v and the fact that air resistance FR increases with speed. The power is eventually
derivable from burning of the fuel inside the engine. We also say that
P
energy consumed
time
for the power consumption that generates the traction force by spinning the tires.
When the car is accelerating, the tractive force is more than air resistance, and the power
demand increases.
13
When the car is going uphill at a constant speed, the tractive force must balance both air
resistance and the component of its weight down the road:
FS  FR  mg sin 
FS
v
FN
FR

mg
In this case the power consumption is
P  FR  mg sin  v
and is higher than for the level road if v is the same. On the other hand, if P remains
constant as the car starts to climb uphill, its speed will be reduced.
14
13. Impulse and Momentum
The concept of momentum
Consider an object of mass m acted on by a single force F. We write Newton’s second
law in the form


F  ma


where F is the average force and a is the average acceleration over an time interval t .
The equation can be transformed as follows:








p f  pi
v mv f  vi  mv f  mvi
F m



t
t
t
t
What has happened is that we have introduced a quantity


p  mv
called the momentum of the object, and shown that the difference between the final


momentum p f and the initial momentum p i enters into the equation. We can also write
 p
F
t
or,
 
p  F t

The product Ft is called impulse. The last equation states that the change of
momentum of an object is equal to the impulse it has received, and is known as the
impulse-momentum theorem.
The unit of momentum in the metric system is kg  m / s .
Note that momentum is a vector in the same direction as velocity.
The impulse momentum theorem is useful in analyzing impact phenomena, in which an
object experiences a force for a very short time duration, such as when a tennis ball is hit
by a racket, or during a collision. For these phenomena, we have good knowledge of the
state of motion of the object before and after impact, but what happens during the short
interval of impact itself is complicated. The impulse momentum theorem allows us to
estimate the average force experienced by the object during impact, as the following
examples show.
15
Example: A 0.05 kg bullet traveling at 1,400 m/s is shot into a wooden target of thickness
0.1 m and emerges from it with the reduced velocity of 1,000m/s. Find the direction and
magnitude of the average force experienced by the bullet.
Solution: First, we can find the duration of impact from the thickness of the target and the
average velocity
v
1400  1000
 1200m / s
2
t 
0.1
 8.3  10 5 s
1200

We can find  p from an examination of the diagram, which shows its construction
pi
pf
Δp
x
The magnitudes of the initial and final momentum are

pi  0.05  1400  70kg  m / s

p f  0.05  1000  50kg  m / s

We see from the diagram that  p is opposite to the velocity of the bullet, and has a
magnitude of

p  70  50  20kg  m / s
The magnitude of the force is


p
20
F 

 2.4  10 5 N
t 8.3  10 5
Thus very strong force occurs during impact.
Another method of doing the problem makes use the components along the x-axis as
shown.
16
x  component of initial momentum pi  70kg  m / s
x  component of final momentum p f  50kg  m / s

 x  component of p p  50  70  20kg  m / s

p
 20
x  component of F
F

 2.4  10 5 N
5
t 8.3  10
The negative sign indicates that the force is opposite to direction of the x-axis.
Example: A 0.050 kg tennis ball traveling at 50m/s hits a racket and is returned with a
speed of 30m/s. If the duration of impact is 1ms, what is the average force on the racket?

Solution The difference in momentum  p is as shown:
pf
pi
Δp

from which we see that p  p f  pi  0.05  50  0.05  30  0.4kg  m / s
The magnitude of the force on the ball is

p
0.4
F 

 400 N
t 1  10 1
In the component method, choosing the x-axis to point in the incoming direction of the
ball, the x-component of the force on the ball is
F
p p f  pi 0.05   50  0.05   30


 400 N
t
t
1  10 3
and is thus opposite to the incoming direction. The force on the racket is equal in
magnitude but opposite in direction.
Forces arising from a continuous jet of water directed onto a surface can also be found
from momentum considerations as the following example shows.
Example: Rain pours down continuously at a speed of 20m/s on the rooftop of a car and
at a rate of 0.3 kg/s. Find the force on the rooftop.
Solution: We use a device of introducing a time interval t to help reasoning. The time
interval t will eventually drop out of the final answer.
The mass of water fallen on the rooftop during the interval t is
 t
m  m
17
where m  mass fallen on rooftop per sec  0.3 kg / s
The initial momentum carried by this mass of water is downward and equals to
pi  m  v  m vt
in magnitude where v  20m / s is the speed of the rain drops. The final momentum is
zero because when the rain drops splashes off the rooftop, its speed is reduced to zero.
Therefore the magnitude of the force on the mass of rain being considered is
F
p m vt

 m v  0.3  20  6 N
t
t
The force on the rooftop is equal in magnitude and directed downward.
We can say that the downward momentum carried by the raindrops is destroyed by the
upward force due to the rooftop, which then experiences an equal force in the opposite
direction.
18
14 Conservation of Momentum
Collisions form a class of impact phenomena because the durations are short and the
forces involved are large. Consider the collision between two objects, labeled 1 and 2.
Let their initial momenta (before collision) be p1i and p2i and their final momenta (after
collision) be p1f and p2f.
P1
P1i
f
P2i
P2
f
Each object changes its momentum because of the impulse it receives during the
collision. Applying the impulse-momentum theorem to each object gives


p1  F1 t


p 2  F2 t


where the F1 is the force on object 1 due to object 2, and F2 is the force on object 2 due to


object 1. Accordingly to Newton’s third law, F1 and F2 are equal and opposite:
 
F1  F2  0
Therefore


p1  p2  0
Or,
 
 p1  p2   0
 
The sum p1  p2 is the total momentum of the system of two objects. The sum is not
changed by collision. The statement that the total momentum before collision is equal to
the total momentum after collision is known as conservation of momentum. The equation
expressing this in the case of two body colliding and flying afterward is




p1i  p 2i  p1 f  p 2 f
19
The conservation of momentum holds in the general case of any number of objects
involved in a collision with any number coming out of the collision. .In this case, we
write


pi  p f
In component form, the above is equivalent to the following equations
for the x and y components respectively:
pix  p fx
piy  p fy
1  2 collisions
Example: An 80-kg adult and a 20-kg child stand together in an ice skating ring. After
pushing each other away, the child is observed to travel with speed 6m/s. What is the
speed of the adult and in what direction?
Solution: Choose the x-axis to point in the direction of the child’s velocity.
x-component of the total momentum before collision =0 because neither object has
velocity
x-component of the total momentum after collision = 80  va  20  6 where va is the xcomponent of the velocity of the adult. Conservation of momentum implies
80va  120  0
 va  
120
 1.5m / s
80
Thus the adult moves with speed 1.5m/s in the opposite direction to the child.
When a cannon ball shoots out from the barrel of a gun with high speed, it carries away
momentum in the forward direction. By conservation of momentum, the gun barrel
acquires momentum backward and a smaller velocity because it has much more mass
than the cannon ball. The motion of the gun barrel is called a recoil.
An astronaut severed from his spaceship and floating freely in space can return to the
spaceship by throwing his tool box in the direction opposite to the space ship. He will
acquire a velocity toward the space ship because of recoil.
20
Example: The breakup of a nucleus in flight. The reaction
U 238  He 4  Th 234
describes the decay of a uranium nucleus U of mass number 238 to a helium nucleus
(mass number 4) and a thorium nucleus (mass number 234). The velocity of the uranium
necleus is 5  10 4 m / s . The velocity of the helium nucleus is 2  10 6 m / s in the opposite
direction. Find the velocity of the thorium nucleus.
Solution: Choose the x-axis to point in the direction of the velocity of the uranium
nucleus.
Apply momentum afterward  momentum before


234vTh  4   2  10 6  238  5  10 4
234vTh  8  10 6  11.90  10 6
 vTh 
11.9  10 6  8  10 6
 1.7  10 4 m / s
234
A pictorial representation of the decay process is as shown:
2  1 Collisions
Example A 1500-kg car travelling at 40m/s runs into a stationary 1000-kg car. The two
become stuck together. Find the velocity of the wreck.
21
Solution: Choose x-axis to point in the direction of the moving car. Let
v be the velocity of the wreckage.
Apply momentum afterward  momentum before
1500  1000 v  1500  40  1000  0
v 
1500  40
 24m / s
2500
Example: A 1500-kg car travelling at 30m/s collides head on with a 1000-kg car
travelling at 10m/s in the opposite direction. The two become stuck together. Find the
velocity of the wreck.
Solution: With the same choice of x-axis, the conservation of momentum gives
1500  1000v  1500  30  1000   10
v 
35000
 14m / s
2500
Comparing the last two examples, we see that the result of the last one can be obtained
from the previous one by adding a velocity of -10m/s to the velocity of every car before
and after the collision. In fact, the two examples describe the same accident as observed
by two pedestrians. One is stationary, and the other is running with speed 10m/s in the
same direction as the velocity of the 1500-kg car. The last pedestrian would see the 1500kg travelling at 30m/s and the 1000-kg car at 10m/s in the opposite direction.
This demonstrates that the conservation of momentum can be used by different observers.
Although momentum is conserved during a collision, kinetic energy usually is not.
Example: What is the change of kinetic energy in the previous example involving a
moving and a stationary car?
Solution: Total Kinetic energy before collision K i 
Total Kinetic energy after collision K f 
1
 1500  40 2  1.20  10 6 J
2
1
 2500  24 2  0.72  10 6 J
2
 K  K f  K i  0.48 106 J  4.8 105 J
22
Thus there is a loss of kinetic energy, which becomes heat.
Example: A object of mass m traveling with speed v collides with a stationary target of
mass 2m . After collision, the light object stops while the heavier one travels with speed
v 2 . Is momentum conserved? Is mechanical energy conserved?
Solution: Initial momentum = mv
Final momentum = 2m  
v
 mv
2
Therefore momentum is conserved.
1
Initial kinetic energy = mv 2
2
2
Final kinetic energy =
1
1
v
 2m      mv 2
2
4
2
Therefore kinetic energy is not conserved.
Non-conservation of mechanical energy in a collision is the norm. Such collisions are
called inelastic collisions. A collision which conserves mechanical energy is called an
elastic collision.
For a two-body elastic collision in one dimension, the velocities of the two after
collision can be determined. Let
mass of the incoming object (the projectile)  m1
mass of the target  m2
velocity of projectile before collision  v1i
velocity of target before collision  0 (stationary target)
velocity of projectile after collision  v1 f
velocity of target after collision  v 2 f
From conservation of momentum,
m1v1 f  m2 v 2 f  m1v1i
From conservation of mechanical energy
1
1
1
m1v12f  m2 v 22 f  m1v12i
2
2
2
These are two equations in the two unknowns v1 f and v 2 f .
23
The solution of these two equations proceeds as follows. If you are not interested, you
can skip this part.
Solution: We shall eliminate v1 f from the equations to end up with an equation for v 2 f .
Define the mass ratio r 
m2
m1
Dividing each equation by m1 yields
v1 f  rv 2 f  v1i
v12f  rv22 f  v12i
From the first equation
v1 f  v1i  rv 2 f
Substitute into the second equation:
v
1i
 rv 2 f

2
 rv 22 f  v12i
v12i  2rv1i v2 f  r 2 v22 f  rv22 f  v12i
r
2

 r v22 f  2rv1i v2 f  0


v 2 f r  1v 2 f  2v1i  0
Thus there are two possibilities:
v2 f  0
or v 2 f 
2
v1i
1 r
The first is uninteresting because it leads to
v1 f  v1i  rv 2 f  v1i
which means that collision does not take place.
Thus we take the second choice, which also gives
24
v1 f  v1i  rv 2 f  v1i 
2r
1 r
v1i 
v1i
1 r
1 r
Replacing r by m2 m1 yields the final form of the solution:
v1 f 
m1  m2
v1i
m1  m2
v2 f 
2m1
v1i
m1  m2
Special case 1: Light projectile and heavy target m1  m2  . In thus case we can neglect
m1 in comparison with m2 .
 v1 f 
 m2
v1i  v1i
m2
v2 f 
2m1
v1i
m2
Thus the projectile is reflected off the target with a speed essentially equal to its incoming
speed, while the target moves very slowly after collisions.
Special Case 2: Equal mass collision m1  m2  .
v1 f  0
v2 f 
2m1
v1i  v1i
m1  m1
The velocity of the projectile is transferred to the target at 100%.
Special Case 3: Heavy projectile and light target m1  m2  . Neglecting m2 in
comparison with m1 yields
v1 f 
m1
v1i  v1i
m1
v2 f 
2m1
v1i  2v1i
m1
Thus the projectile velocity is essentially unchanged, while the target has twice the initial
velocity of the projectile.
The Ballistic Pendulum is an arrangement that can be used to measure the speed of a
bullet. The bullet is shot into a wooden block hanging from a support, and becomes
embedded in the block. Subsequently the block with the bullet in it swings upward and
reach a highest position. By measuring the rise in height, we can deduce the speed of the
bullet, provided the masses of the bullet and the block are known. The diagram is a
sketch of the situation.
25
m
v
u
M
h
M+
m
Let
m  mass of bullet
M  mass of block
v  speed of bullet
u  speed of block  bullet right after collision
h  max imum height of block above rest position
We can apply conservation of momentum to the impact between the bullet and the block:
m  M u  mv
which can be solved to give
v
mM
M
u
u
m
m
as the mass of the bullet is usually much smaller than that of the block.
We then apply conservation of mechanical energy to the rise of the block+bullet system
to its maximum height when the speed is zero:
1
m  M u 2  m  M gh
2
which gives
u  2 gh
v 
M
M
u
m
m
2 gh
26
For example, with m  0.1kg M  2kg h  0.05m we find
v
2
2  10  0.05  20m / s
0.1
For collisions in two dimension, we need to apply conservation of momentum separately
for the x and the y components as the following example shows.
Example: A 1500-kg car traveling at 40m/s due east collides with a 1000-kg car traveling
at 50m/s due north. The two become stuck together. Find the speed and direction of the
wreckage.
y
40m/
s
v
x
1500k
g
50m/
s
2500k
g
1000k
g
Solution: Choosing x and y axes as shown.
The x-component of momentum conservation gives
2500v x  1500  40
 v x  24m / s
The y-component of momentum conservation gives
2500v y  1000  50
 v y  20m / s
The magnitude of v is

v  24 2  20 2  31m / s
  tan 1
20
 40 
24
The direction is 40º N of E.
27
Motion of a System of Particles.
For a system of particles, we can divide up the forces on each particle into external and
internal forces. Internal forces on each particle are due to other particles in the system,
and they occur in pairs of equal magnitude and opposite directions, acting on differing
particles, according to Newton’s third law. There we expect the sum of all internal forces
on all the particles to cancel out in pairs, leading to

Fint  0
For each particle, Newton’s second law states


p 
 Fext  Fint
t
Summing over all particles,





p
  Fext   Fint   Fext
t
If we introduce the total momentum of the system


Pp
the above can be written as


P
  Fext
t
The total momentum is related to a concept called the center of mass. In one dimension,
if the coordinates of the point masses m1 , m2 , are x1 , x2 ,, the coordinate of the center
of mass is obtained from
x
m1 x1  m2 x2  

m1  m2  
 mx
m
In two dimension, we add to the above equation the y-coordinate of the center of mass
y
 my
m
The position vector of the center of mass is therefore
28

r

 mr
m
Or,


Mr   mr
where M   m is the total mass of the system.
When positions of the particles change, so does the center of mass. The relation between
the displacements are


Mr   mr
Upon diving by t , the above equation gives


 
Mv   mv   p  P
Thus, the total momentum is equal to the total mass times the velocity of the center of
mass. The equation for the motion of the center of mass can be written as



v P
M

  Fext
t
t
It is significant that we require no knowledge of internal forces when determining the
acceleration of the center of mass. A rigid box can be considered as a system of particles
if we imagine mentally dividing it into small bits. If you pull it with a string, an external
force is applied. To find the rate of acceleration of the center of mass, we need only know
the mass of the block and the pull force. We have been using this fact in our course all
along.
A system with no external forces is called an isolated system. For such system, we see

P
that
0
t

 P  0
The total momentum is conserved for an isolated system. The center of mass of an
isolated system travels with constant velocity.
29
15. Kinematics of Rotation
The radian measure of an angle is defined by the formula

s
r
where r is the radius of a circle centered at the apex of the angle, and s is the arc length
on the circle included between the two sides of the angle. The value of  so obtained is
independent of the size of the circle.
s

r
For convenience in describing rotational motion, we can understand an angle in the
generalized sense in the following way. Imagine an x-axis drawn fixed on a paper, and a
rotating line segment drawn with one end fixed at the origin. As the segment rotates, it
opens up an angle with the x-axis. The angle is considered positive if the sense of rotation
is counter clockwise (ccw), and negative if clockwise (cw).
When a complete ccw rotation is made, the angle is 360  , the radian measure of this
angle is
circumference 2r

 2
radiu
r
Thus 2 radian equals 360º. Therefore
1 rad 
360 
 57.3
2
Starting with an angle  , rotating the segment through one complete revolution brings it
back to the starting position. The same is true after multiple revolutions in one direction
or the other. Therefore the angles
30
 ,   2 ,   4 , ,  2 ,   4 ,
all represent the same position.
Apparent size of an object.
The apparent size of an object depends on the solid angle subtended at the eye by light
rays emanating from the edges of the object. These light rays from a pencil-like
configuration with the tip at the eye. In the case of a circular disk, the apparent size is
governed by a planar angle subtended at the eye by the two rays coming from the ends
of a diameter.
r
d
s
Suppose d is the diameter, and the object is quite far away so that its distance is equal to
the radius of a circle centered at the eye but include the diameter of the object as a chord.
Let  denote the angle subtended at the eye by the object. Note that arc length s is
approximately equal to the diameter d if the distance is large. Thus we have

s d

r r
If we apply this to the moon,
d  diameter of the moon  2  1.74  106 m  3.5  10 6 m
r  dis tan ce of the moon  3.84  1011 m
 
3.5  10 6
360
3
3

 0.5
8  9.2  10 rad  9.2  10
3.8  10
2
It turns out that a similar calculation for the apparent size of the sum also yields 0.5,
which explains why we sometimes have total solar eclipse and some times have annular
eclipse, as the distance between the earth and the sun has slight variability in the course
of a year.
The smallest angle an optical instrument can detect is called its resolving power. If the
resolving power of the eye is 1/20 that of the apparent size of the moon, what is the
maximum distance from which a coin of diameter 2cm can be seen as a tiny disk?
31
Solution:  
r
d


1
2
4
 0.5  0.025  0.025 
 4.4  10 rad
20
360
0.02
4  45m
4.4  10
which is a little less than half a football field away.
Rotation of a rigid body In the simplest kind of rotation of a rigid body ( a rod, a disk,
a bike wheel, etc), there is an axis of rotation that remains fixed in space. To keep track
of the position of the rigid body, we can mark off a point P on the body, which then
would move in a circle with the axis of rotation O as the center. We then choose an x-axis
as reference, with its origin at the axis of rotation. The angle  between the line segment
OP and the x-axis can be used to represent the position of the rigid body, just like in the
case of motion of a point object along the x-axis, the coordinate x is used to represent its
position. After a time interval t , the angle changes by the amount   . We define the
average angular velocity by
 

t
and the instantaneous angular velocity by
  lim

t
as t  0
which is roughly speaking the instantaneous velocity over a very small time interval.
The unit of angular velocity is rad/s. Another commonly used unit is rpm (revolutions per
minute). The conversion factor is
1 rpm 
2 rad
 0.105 rad / s
60 s
The average and instantaneous angular acceleration are defined by
 

t
  lim

as t  0
t
For motion of constant acceleration,  is a constant. The equations governing the
kinematics of such rotations can be developed in parallel with the kinematics for uniform
acceleration on a straight line using the following correspondence:
32

0
x
v0


   0  t
v
a
v  v0  at
1
2
1 2
at
2
v v
v 0
2
x  vt
   0 t  t 2
 
x  v0 t 
0  
  t
2
 2   02  2
v 2  v02  2ax
Example: A flywheel is initially rotating at 2500rpm. It is slowing down at a rate of 50
rad/s. How long does it take to stop and how many complete revolutions it has gone
through before stopping?
Solution: We list the known quantities as follows, considering the rotation to be ccw:
 0  2500rpm  2500 
2
rad / s  262rad / s
60
  0 stopping 
  50 rad / s2
t 
   0 0  262

 5.2s

50
  t 
  0
2
t
0  262
 5.2  680rad
2
Number of revolutions =

680

 108
2
2
Tangential velocity
Any point P on a rigid body rotating about an axis O travels in a circle of radius r with O
as center. If vT is the magnitude of the instantaneous velocity of the point P, we expect
vT to increase with r . For example, if you spin a meter stick about one end, the farther a
point is from this end, the higher is the speed of the point. To derive the quantitative
relationship, we choose a very small t ., during which the point P moves through a small
arc of length s , the the line segment OP rotates through a small angle   . From the
relation
33
s  r
we divide both sides by t to obtain
s

r
t
t
In the limit t  0 , this becomes
vT  r
where  is the instantaneous angular velocity of the rigid body.
Transfer of Angular Velocity
One way to step up or step down angular velocity can be illustrated by two cogged
wheels in a bike connected by a chain. Since an equal of length of the chain is released
from one wheel and taken up by the other, the linear speed of the release point on the first
wheel is equal to the linear speed of uptake on the second wheel. If the radii of the two
wheels are denoted by r1 and r2 , and the magnitude of their angular velocities are 1 and
 2 , and further the common linear speed is v , we have
v  r11
v  r22
r11  r22
r2
Thus the smaller wheel rotates with the higher angular velocity. In this case both wheels
rotate in the same sense.
When the two wheels are in contact at one point, so that their teeth are locked into each
other, the contact point between the wheels share the same tangential speed. The above
34
derivation still applies. Except that now the senses of rotation of the two wheels are
opposite.
Example: CD playing. A laser beam is used to read the information stored in concentric
circular tracks of a compact disk. The information is read at the rate of 1.3 m/s, which
means a length of record of 1.3 m is read per second. As the reading progresses, the laser
beam is shifted from the inside track to the outside track. The radius of the innermost
track is 23mm, and that of the outermost track is 58mm. The total playing time is
75minute and 33 seconds, which equals 4473s. From these data we can calculate
the angular velocity at the beginning
1 
v
1.3
60

rpm  540rpm
3  56.5rad / s  56.5 
r1 23  10
2
the angular velocity at the end
2 
v
1.3
60

rpm  210rpm
3  22.4rad / s  22.4 
r2 58  10
2
Thus the disk is slowing down in its rotation as the music is played. The angular
acceleration is

 2  1

t
22.4  56.5
3
2
 7.6  10 rad / s
4473
The total angle rotated through is

1   2
2
5
t  1.76  10 rad
The number of complete revolutions is

 28,000
2
The total length of record read is
vt  1.3  4473  5800m  3.6mi
Rolling Motion When a bicycle wheel rolls on the ground to carry the biker forward, the
axis of rotation is not fixed in space, but instead moves forward. If there is no slipping,
the curved length on the wheel unfolds into the horizontal distance on the ground. The
diagram shows the position of the wheel at two times, together with points P and Q
marked off on the wheel. If s is the arc length PQ,  is the angle between OP and OQ,
and x is the forward distance traveled by the point O, the condition of no slipping is
35
x  s
Since s  r , we have
x  r
relating the forward distance to the angle of rotation, Dividing both sides by t leads to
vO  r
relating the forward velocity of the point O to the angular velocity of rotation.
P
O

Q
Δs
O
Q
P
Δx
Example: What is the angular velocity of the wheels of a car traveling at 60mph, if the
diameter of each wheel is 80cm? How many complete turns does each wheel make after
the car has traveled 1 mile?
Solution: Part 1: vO  60mph  26.8m / s
0. 8
r
 0. 4 m
2
v
26.8
  O 
 67rad / s
r
0.4
Part 2: x  1mi  1.6  10 3 m
 
x 1.6  10 3

 4  10 3 rad
r
0.4
number of revolutions 

 637
2
36